SN2 organic question

[student1982]

Why is the product an ether rather than just another OH added to the Cl end. --> HO-CH2-CH2-CH2-C2-OH



HO-CH3-CH3-CH3-CH2-Cl + NaOH -->



CH2-CH2-CH2
\.........../
....CH2-O

Also, anybody know of a web page that can hold clip pictures so you can show a picture when you post on here. It would be mouch easier to post a picture rather than try to draw it out with little lines using a keyboard. thanks

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[chessxwizard]

So this question makes you think a little.

Remember in organic chemistry how 5-membered rings formed readily?

Well, in this case, NaOH dissociates into Na+, which is not a nucleophile, and OH-, which is.

OH- has the opportunity to attack either the hydrogen of the alcohol group at the end of the compound or displace via SN2 the Cl on the other end.

Keep in mind that the terminal alcoholic hydrogen is very acidic, and is readily taken off by the basic OH- produced by the dissociation of sodium hydroxide.

The O- at the end of the molecule, now a good nucleophile, attacks the Cl attached to the other side of the molecule to create an cyclic 5-membered ether, an extremely stable structure.

Hope this helped.

As for drawing pictures and posting them here, chemdraw and photobucket are pretty good tools for organic chemistry discussions. If you don't have chemdraw, microsoft paint will suffice.

 

[student1982]

Thanks a lot. Very good explanation. For whatever reason I thought that H on the alc end is extremely hard to take off, but now I know. Also, will the product be a mixture of the cyclic ether & 1,4-butanediol ?
HO-CH2-CH2-CH2-CH2-OH

I still can't see why that OH- can't just attack the Cl end and be stable?
I mean its not hindered, good leaving group, & OH- is a strong nuc

 

[tastybeef]

You will generally have an intramolecular Sn2 reaction when the product is a 5 or 6-membered ring. The ring structure is stable.

While OH- is a strong nucleophile, it is an even better base, so it will take the hydrogen off the hydroxy group on your starting material first.